Waterbury Insurance Company wants to study the relationship between the amount o
ID: 3339757 • Letter: W
Question
Waterbury Insurance Company wants to study the relationship between the amount of fire damage and the distance between the burning house and the nearest fire station. This information will be used in setting rates for insurance coverage. For a sample of 30 claims for the last year, the director of the actuarial department determined the distance from the fire station (X) and the amount of fire damage, in thousands of dollars (Y). The MegaStat output is reported below.
Is there a direct or indirect relationship between the distance from the fire station and the amount of fire damage?
How much damage would you estimate (in dollars) for a fire 7 miles from the nearest fire station? (Round your answer to the nearest dollar amount.)
Fill in the blank below. (Round your answer to 1 decimal place.)
State the decision rule for .01 significance level: H0 : = 0; H1 : 0. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)
Waterbury Insurance Company wants to study the relationship between the amount of fire damage and the distance between the burning house and the nearest fire station. This information will be used in setting rates for insurance coverage. For a sample of 30 claims for the last year, the director of the actuarial department determined the distance from the fire station (X) and the amount of fire damage, in thousands of dollars (Y). The MegaStat output is reported below.
Explanation / Answer
1a) y= 13.7523+ 6.3449X
2a) Direct relationship between distance and damage .
3) Y= 13.7523+6.3449*7= 58.17 dollars
4) coefficient of determination from the table = Coefficient of Determination = r2 = SS(Regression) / SS(Total)
1865.5782/3186.0716= 0.585
5) 58.5% of variation explained.
6) Correlation coefficient is sqrt(0.585)= 0.764
Sign of correlation coefficient is positive , sign of slope is positive
7) Decision rule is reject null hypothesis if calculated value of t is greater than critical value of t with 28 degree of freedom at .01 significance level 2.763
8) test statistic= r*sqrt(n-2)/sqrt(1-r^2)= (0.764* 5.29)/0.644
=6.27
t= 6.27
Calculated value of t is greater than critical value of t with 28 degrees of freedom at .01 level of significance. We may conclude that we will reject null hypothesis..
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